By Jerrold Bebernes

ISBN-10: 0387971041

ISBN-13: 9780387971049

ISBN-10: 146124546X

ISBN-13: 9781461245469

ISBN-10: 146128872X

ISBN-13: 9781461288725

This monograph developed during the last 5 years. It had its foundation as a collection of lecture notes ready for the 9th summer time tuition of Mathematical Physics held at Ravello, Italy, in 1984 and used to be extra subtle in seminars and lectures given basically on the college of Colorado. the cloth provided is the fabricated from a unmarried mathematical query raised by way of Dave Kassoy over ten years in the past. this query and its partial solution resulted in a profitable, fascinating, virtually particular interdisciplinary col­ laborative clinical attempt. The mathematical versions defined are usually occasions deceptively uncomplicated in visual appeal. yet they convey a mathematical richness and sweetness that belies that simplicity and affirms their actual importance. The mathe­ matical instruments required to unravel a few of the difficulties raised are various, and no systematic try out is made to provide the mandatory mathematical heritage. The unifying topic of the monograph is the set of versions themselves. This monograph may by no means have come to fruition with out the enthu­ siasm and force of Dave Eberly-a former scholar, now collaborator and coauthor-and with no a number of major breakthroughs in our comprehend­ ing of the phenomena of blowup or thermal runaway which sure versions mentioned own. A collaborator and previous scholar who has made major contribu­ tions all through is Alberto Bressan. there are various different collaborators­ William Troy, Watson Fulks, Andrew Lacey, Klaus Schmitt-and former students-Paul Talaga and Richard Ely-who has to be stated and thanked.

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Extra info for Mathematical Problems from Combustion Theory

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Proof. Let R E (0, 1) be the first value of r such that u"(r) = 0. Define m := u'(R); then u(R) =In [ -mg~- 1 )]. 17), let r = sR and v(s) = u(r)- u(R). 28) (n-1)/3-8=0 where 8 = -(n- 1)mR > 0 and /3 = -mR > 0. 26, there is a unique pair (8, /3) satisfying (n- 1)/3- 8 = 0 and a corresponding unique solution v(s). Thus, v"(1) = 0. Since u"(r) < 0 on 0 ~ r < R, v"(s) < 0 for 0 ~ s < 1. Suppose there as a value P E (R, 1] such that u"(R) = u"(P) = 0. Set f = u'(P); then u(P) = In [ -t~~- 1 )]. Make a change of variables r = sP and v(s) = u(r)- u(P).

O and, for>. t). 11) 30 2. L = A1 . L > A1 . ; E 0. 14 applies. L-e). L). 13), 0 = u(x) ~ u(x~') which is a contradiction to u > 0 on 0. Thus, x E T~' and x = x~'. Fork sufficiently large, the line segment joining xik and x::k is in 0. 12) and the Mean Value Theorem, there is a Yik such that Uxt (Yik) ~ 0. L. l > A1 is incorrect. 10) is valid for all A E (AI, A0 ). 16 If Uxt (x) = 0 for some x E 0 in TAt and n T>'l, then u is symmetric Proof. ). This implies that u is symmetric relative u(x) to TAt' Since u(x) > 0 in E(Al) and u = 0 on 80, we conclude that 0 = E(Al) U E'(Al) U [TAt n OJ.

29}. The proof uses perturbation arguments and is quite involved and technical. 8. The value e is positive and chosen sufficiently close to 0. 8 are called small-small solutions. Those solutions represented by the broken line (- -) are called large-small solutions. Those solutions represented by dotted line (- · -) are called large solutions. 8 may not be exactly correct. The branch representing large-small solutions is arbitrarily close to the small-small solution branch for c > 0 but small. 4 on solution profiles.

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